binomial and poisson probablity distribution for DSML

Discrete
Probability
Distributions
Chapter 4

Random Variables
A random variable x represents a numerical value
associated with each outcome of a probability distribution.
A random variable is discrete if it has a finite or countable
number of possible outcomes that can be listed.
x
0 2 4 6 8 10
A random variable is continuous if it has an uncountable
number or possible outcomes, represented by the intervals on
a number line.
x
0 2 4 6 8 10
Larson & Farber, Elementary Statistics: Picturing the World, 3e 2

Random Variables
Example:
Decide if the random variable x is discrete or continuous.
a) The distance your car travels on a tank of gas
The distance your car travels is a continuous random
variable because it is a measurement that cannot be
counted. (All measurements are
continuous random variables.)
b) The number of students in a statistics class
The number of students is a discrete random
variable because it can be counted.

Discrete
Probability
Distributions
A discrete probability distribution lists each possible
value the random variable can assume, together with its
probability. A probability distribution must satisfy the
following conditions.
In Words
1. The probability of each value of
the discrete random variable is
between 0 and 1, inclusive.
In Symbols
0  P (x)  1
2. The sum of all the
probabilities is 1.
ΣP (x) = 1

Constructing a Discrete
Probability Distribution
Guidelines
Let x be a discrete random variable with possible
outcomes x1, x2, … , xn.
1. Make a frequency distribution for the possible
outcomes.
2. Find the sum of the frequencies.
3. Find the probability of each possible outcome by
dividing its frequency by the sum of the frequencies.
4. Check that each probability is between 0 and 1 and
that the sum is 1.

2
Example:
The spinner below is divided into two sections. The
probability of landing on the 1 is 0.25. The probability of
landing on the 2 is 0.75. Let x be the number the spinner
lands on. Construct a probability distribution for the
random variable x.
1
x P
( x)
1 0.25
2 0.75
Each probability is
between 0 and 1.
The sum of the probabilities is 1.

Example:
The spinner below is spun two times. The probability of
landing on the 1 is 0.25. The probability of landing on the 2
is 0.75. Let x be the sum of the two spins. Construct a
probability distribution for the random variable x.
2
1
The possible sums are 2, 3, and 4.
P (sum of 2) = 0.25  0.25 = 0.0625
Spin a 1 on “and”
the first spin.
Spin a 1 on the
second spin.
Continued.
7
Larson & Farber, Elementary Statistics: Picturing the World, 3e

2
Example continued :
P (sum of 3) = 0.25 
0.75 = 0.1875
1
Spin a 2 on the
second spin.
the first spin.
“or”
P (sum of 3) = 0.75  0.25 = 0.1875
“and” Spin a 1 on the
second spin.
3
4
P (x)
Sum of
spins,
x2
Continued.
0.0625
0.375
Spin a 2 on
the first spin.
0.1875 + 0.1875

Example continued :
2
1 P (sum of 4) = 0.75  0.75 = 0.5625
the first spin.
Spin a 2 on the
second spin.
Sum of
spins,
P (x)
x2 0.0625
3 0.375
4 0.5625
Each probability is between 0
and 1, and the sum of the
probabilities is 1.

Graphing a Discrete
Probability
Distribution
Example:
Graph the following probability distribution using a histogram.
Sum of
spins,
P (x)
x2 0.0625
3 0.375
4 0.5625
Sum of Two Spins
x
Probability
0.2
0.1
0
0.6
0.5
0.4
0.3
2 4
3
Sum
P(x)

Mean
x P (x) xP (x)
2 0.0625 2(0.0625) = 0.125
3 0.375 3(0.375) = 1.125
4 0.5625 4(0.5625) = 2.25
The mean of a discrete random variable is given by
μ = ΣxP(x).
Each value of x is multiplied by its corresponding
probability and the products are added.
Example:
Find the mean of the probability distribution for the sum of
the two spins.
ΣxP(x) = 3.5
The mean for the
two spins is 3.5.

Variance
approximately 0.376
x P (x) x – μ
(x –
μ)2
P (x)(x –
μ)2
2 0.0625 –1.5 2.25  0.141
3 0.375 –0.5 0.25  0.094
4 0.5625 0.5 0.25  0.141
The variance of a discrete random variable is given by
2 = (
Σ x – μ)2P (x).
Example:
Find the variance of the probability distribution for the
sum of the two spins. The mean is 3.5.
ΣP(x)(x – 2)2
 0.376
The variance for the
two spins is

Standard Deviation
The standard deviation of a discrete random variable is
given by
σ = σ 2
.
x P (x) x – μ
(x –
μ)2
P (x)(x –
μ)2
2 0.0625 –1.5 2.25 0.141
3 0.375 –0.5 0.25 0.094
4 0.5625 0.5 0.25 0.141
Example:
Find the standard deviation of the probability distribution
for the sum of the two spins. The variance is 0.376.
σ 
σ 2

0.376  0.613
Most of the sums

Binomial Experiments
A binomial experiment is a probability experiment
that satisfies the following conditions.
1. The experiment is repeated for a fixed number of
trials, where each trial is independent of other trials.
2. There are only two possible outcomes of interest for
each trial. The outcomes can be classified as a
success
(S) or as a failure (F).
3. The probability of a success P (S) is the same for each
trial.
4. The random variable x counts the number of
successful trials.

Notation for
Binomial
Experiments
Symbol
n
p = P (S)
q = P (F)
x
Description
The number of times a trial is repeated.
The probability of success in a single trial.
The probability of failure in a single trial.
(q = 1 – p)
The random variable represents a count of
the number of successes in n trials:
x = 0, 1, 2, 3, … , n.

Example:
Decide whether the experiment is a binomial experiment.
If it is, specify the values of n, p, and q, and list the possible
values of the random variable x. If it is not a binomial
experiment, explain why.
• You randomly select a card from a deck of cards, and
note if the card is an Ace. You then put the card
back and repeat this process 8 times.
This is a binomial experiment. Each of the 8 selections
represent an independent trial because the card is
replaced before the next one is drawn. There are only two
possible outcomes: either the card is an Ace or not.
52 13
p  4  1
n  8 q  1  1  12 x  0,1,2,3,4,5,6,7,8
13 13

Example:
Decide whether the experiment is a binomial experiment.
If it is, specify the values of n, p, and q, and list the possible
values of the random variable x. If it is not a binomial
experiment, explain why.
• You roll a die 10 times and note the number the die
lands on.
This is not a binomial experiment. While each
trial (roll) is independent, there are more than two
possible outcomes: 1, 2, 3, 4, 5, and 6.

Binomial Probability
Formula
Example:
A bag contains 10 chips. 3 of the chips are red, 5 of the chips are
white, and 2 of the chips are blue. Three chips are selected, with
replacement. Find the probability that you select exactly one red
chip.
In a binomial experiment, the probability of exactly x
successes in n trials is
n !
px
qn x
.
P (x )  n C x px
qn x

(n  x )!x !
10
p = the probability of selecting a red chip  3  0.3
q = 1 – p = 0.7
n = 3
x = 1
P (1)  3C1(0.3)1
(0.7)2
 3(0.3)(0.49)
 0.441

Binomial
Probability
Distribution
Example:
A bag contains 10 chips. 3 of the chips are red, 5 of the chips are
white, and 2 of the chips are blue. Four chips are selected, with
replacement. Create a probability distribution for the number of red
chips selected.
3
p = the probability of selecting a red chip 
10
 0.3
q = 1 – p = 0.7
n = 4
x = 0, 1, 2, 3, 4
x P (x)
0 0.240
1 0.412
2 0.265
3 0.076
4 0.008
The binomial
probability
formula is used
to find each
probability.

Finding Probabilities
x P (x)
0 0.24
1 0.412
2 0.265
3 0.076
4 0.008
Example:
The following probability distribution represents the probability of
selecting 0, 1, 2, 3, or 4 red chips when 4 chips are selected.
a)
Find the probability of selecting no
more than 3 red chips.
b) Find the probability of selecting at
least 1 red chip.
a.) P (no more than 3) = P (x  3) = P (0) + P (1) + P (2) + P
(3)
= 0.24 + 0.412 + 0.265 + 0.076 = 0.993

Graphing Binomial
Probabilities
x P (x)
0 0.24
1 0.412
2 0.265
3 0.076
4 0.008
Selecting Red
Chips
0.3
x
Probability
0.2
0.1
0
0.5
0.4
0 4
1 2
3
Number of red chips
Example:
The following probability distribution represents the probability of
selecting 0, 1, 2, 3, or 4 red chips when 4 chips are selected. Graph
the distribution using a histogram.
P (x)

Mean, Variance and
Standard Deviation
Population Parameters of a Binomial Distribution
μ  np
σ 2
 npq
σ  npq
Mean:
Variance:
Stand
ard
deviation:
Example :
One out of 5 students at a local college say that they skip breakfast in
the morning. Find the mean, variance and standard deviation if 10
students are randomly selected.
μ  np
 10(0.2)
 2
σ 2
 npq
 (10)(0.2)(0.8)
 1.6
σ  npq

1.6
 1.3
n  10
p  1  0.2
5
q  0.8

Bernoulli Distribution
Bernoulli distribution has only two possible outcomes, namely 1 (success) and
0 (failure), and a single trial. So the random
variable X which has a Bernoulli distribution can take value 1 with the probability
of success, say p, and the value 0 with the
probability of failure, say q or 1-p.
Here, the occurrence of a head denotes success, and the occurrence of a tail
denotes failure.
Probability of getting a head = 0.5 = Probability of getting a tail since there are
only two possible outcomes.
The probability Mass function:

Introduction to the Poisson Distribution
The Poisson probability distribution provides a good model for the probability distribution of the
number of “rare events” that occur randomly in time, distance, or space.
Poisson distribution is for counts—if events happen at a
constant rate over time, the Poisson distribution gives the
probability of X number of events occurring in time T.
Used when no. of trial is very large and chance of success
is small(rare event). E.g
1. No. of air accident in year in India in one year.
2. No. of defective screws per box of 5000 screws

Poisson Distribution
conditions.
P (x )  μ x
e m

The Poisson distribution is a discrete probability distribution
of a random variable x that satisfies the following
1. The experiment consists of counting the number of times
an event, x, occurs in a given interval. The interval can
be an interval of time, area, or volume.
2. The probability of the event occurring is the same for each
interval.
3. The number of occurrences in one interval is independent of
the number of occurrences in other intervals.
The probability of exactly x occurrences in an interval is
x!
where e  2.71818 and μ is the mean number of occurrences.

Poisson Distribution
Example:
The mean number of power outages in the city of Brunswick is 4 per
year. Find the probability that in a given year,
a) there are exactly 3 outages,
b) there are more than 3 outages.
3!
P (3)  43
(2.71828)-4
a.)   4, x  3
 0.195
b.) P (more than 3)
 1  P (x  3)
 1 [P (3)  P (2) +
P (1) + P (0)]
 1  (0.195  0.147 
0.073  0.018)
 0.567

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